it would be fantastic if anyone could help me with the following problem:
I have the integral
$$\operatorname{Im} \left( \int^\infty_0 e^{it} t^{s-1} \mathrm{d} t\right)$$
and I wish to make the variable change $t \to ib$
to give
$$\operatorname{Im} \left( i^{s} \int^\infty_0 e^{-b} b^{s-1} \mathrm{d} b \right) $$
so that I may derive something later on. The only thing is, I am changing the variable and so I change the contour too. Anybody know how to justify that the above holds over the imaginary axis?
Cheers
Just a thought, but you could look at $t$ in the following way
$t = a + ib \qquad t \in [0,\infty) $
where $a=0$. Hence, it follows that $ib$ exists in $[0,\infty)$ of the imaginary plane. And it would also follow that
$\mathrm{d}t = i \cdot \mathrm{d}b $
which you have already made allowances for.